## College Algebra (10th Edition)

If we want to choose $k$ elements out of $n$ regarding the order, not allowing repetition, we can do this in $_{n}P_k=\frac{n!}{(n-k)!}$ ways. The order matters here when choosing the two horses, thus we have to use permutations. We have $8$ horses for $2$ spots, thus $_{8}P_{2}=\frac{8!}{(8-2)!2!}=56$